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2 years ago

What is the position of the earth, moon, and sun during a new moon?

2 answers:

6 0

Answer:When the Moon is between the Earth and the Sun, the bright side of the Moon is facing away from the Earth, and we have a New Moon.

Explanation:

Reika [66]2 years ago

4 0

Answer:

In the case of a New Moon, the Moon is between Earth and the Sun, so the side of the Moon facing the Sun isn't facing us. We can't see any portion of the lit-up Moon during this phase. (When the Sun and Moon exactly line up, as viewed from Earth, we get a special experience known as a solar eclipse

Explanation:

good luck hope this helps

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Many of the stars we see in the night sky are called blue giants. They put out more energy and burn millions of times brighter t

Bumek [7]

Answer:

Because it is closer to earth than the other blue giant stars.

Explanation:

The sun is closer to earth than any other big star which means we see it bigger.

6 0

3 years ago

What type of bond is present between two atoms of tungsten in a light bulb filament?

stiks02 [169]

Answer:

ionic bonding

Explanation:

hope it helps

8 0

3 years ago

A piston is pressurized to 15.5 psi at 405 K. If the piston compresses the air, from 8.98 L to 7.55 L, and the pressure drops to

masya89 [10]

Answer:

239.45 K

Explanation:

Ideal gas law formula is P1V1T2=P2V2T1

Rearrange that to get...

T2=T1P2V2/P1V1

Fill in the values and solve.

7 0

3 years ago

Some magnesium powder was mixed with some copper( II)oxide and heated strongly. there was a victorious reaction producing a lot

-BARSIC- [3]

Reaction:

$\mathrm{Mg \: + \: CuO \rightarrow Cu\: + \: MgO}$

Magnesium is a stronger reducing agent than copper and is thus able to reduce copper(II) oxide.

Products of the reaction: Magnesium oxide and metallic copper.

4 0

3 years ago

Read 2 more answers

Predict the boiling point of water at a pressure of 1.5 atm.

Lina20 [59]

Answer:

100.8 °C

Explanation:

The Clausius-clapeyron equation is:

$ln\frac{P_{1} }{P_{2}} =$ -Δ $\frac{H_{vap}}{r} (\frac{1}{T_{2}}-\frac{1}{T_{1}} )$

Where 'ΔHvap' is the enthalpy of vaporization; 'R' is the molar gas constant (8.314 j/mol); 'T1' is the temperature at the pressure 'P1' and 'T2' is the temperature at the pressure 'P2'

Isolating for T2 gives:

$T_{2}=(\frac{1}{T_{1}} -\frac{Rln\frac{P_{2}}{P_{1}} }{Delta H_{vap}}$

(sorry for 'deltaHvap' I can not input symbols into equations)

thus T2=100.8 °C

7 0

3 years ago